The domain of the function f x -log 1 - 2 x -1-2-log 2-4 x 2-4 x -5 is1-2- -B- - -1-2- -D- -1-2- -

The correct option is A (12,∞)f(x)=log_1/2(x 12)+log_2√(4x^2 4x+5) log_1/2(x 12) is defined if nbsp; x 12 gt;0 i.e., x gt;12, and log_2√(4x^2 4x+5) is de ...

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Standard XII

Mathematics

Domain

The domain of...

Question

The domain of the function $f (x) = {log}_{1 / 2} (x - \frac{1}{2}) + {log}_{2} \sqrt{4 x^{2} - 4 x + 5}$ is

[\frac{1}{2}, \infty)

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(\frac{1}{2}, \infty)

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(- \infty, \infty)

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(- \frac{1}{2}, \infty)

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Solution

The correct option is B $(\frac{1}{2}, \infty)$
$f (x) = {log}_{1 / 2} (x - \frac{1}{2}) + {log}_{2} \sqrt{4 x^{2} - 4 x + 5}$
${log}_{1 / 2} (x - \frac{1}{2})$ is defined if $x - \frac{1}{2} > 0$ i.e., $x > \frac{1}{2},$
and ${log}_{2} \sqrt{4 x^{2} - 4 x + 5}$ is defined if
$4 x^{2} - 4 x + 5 > 0$
$\Rightarrow 4 [{(x - \frac{1}{2})}^{2} + 1] > 0$ which is true for all real $x .$
$∴$ Domain of $f$ is $(\frac{1}{2}, \infty)$

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f (x) = log (1 - x) + \sqrt{x^{2} - 1}

Q. If

{log}_{1 / \sqrt{2}} (1 / \sqrt{8}) = {log}_{2} (4^{x} + 1) . {log}_{2} (4^{x + 1} + 4)

, then

x =

.................

Q. Solve:

2 {log}_{2} ({log}_{2} x) + {log}_{\frac{1}{2}} ({log}_{2} 2 \sqrt{2} x) = 1

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The domain of the function f(x)=log1/2(x−12)+log2√4x2−4x+5 is

The correct option is B (12,∞)f(x)=log1/2(x−12)+log2√4x2−4x+5 log1/2(x−12) is defined if x−12>0 i.e., x>12, and log2√4x2−4x+5 is defined if 4x2−4x+5>0 ⇒4[(x−12)2+1]>0 which is true for all real x. ∴ Domain of f is (12,∞)

The domain of the function $f (x) = {log}_{1 / 2} (x - \frac{1}{2}) + {log}_{2} \sqrt{4 x^{2} - 4 x + 5}$ is